Homework(12(Solutions(!1. !Modeling!bear!weight!(a) Observational!study!with!convenience!sample.!!Cannot!infer!causation.!!Can!only!generalize!to!this!set!of!bears.!(b) µ(Y | Xi, X2, X3) =!0+!1X1+!2X2+!3X3!where!Y!=!weight!(lbs),!X1!=!chest!circumference!(in),!X2!=!length!of!head!(in),!X3!=!neck!circumference!(in)!(c) Y ~ N (!0+!1X1+!2X2+!3X3,"2)!!(normality;!mean!linear!function!of!X1,!X2,!X3;!homogeneous!variance!for!given!set!of!X’s)!and!observations!are!independent!(d) 332211032ˆˆˆˆ),,|(ˆXXXXXXYiββββµ+++==!–264.79!+!9.36!X1!–!0.10!X2!+!5.83!X3!!!!!!!!!!!(These!estimates!are!rounded,!but!you!want!to!use!the!full!set!of!decimals!for!calculations.)!(e) 56.3198383.995ˆ=== MSEσ!(f) Ho:!β1'='β2'='β3!=!0!!vs!!Ha:!at!least!one!βj!≠!0!! F!=!246.49!with!!df1!=!3,!df2!=!50,!pUvalue!<!0.0001!! There!is!strong!evidence!that!mean!weight!µ(Y|X1,'X2,'X3)!depends!linearly!on!one!or!more!of!chest!circumference,!head!length,!and!neck!circumference!(i.e.,!convincing!evidence!that!not!all!βj!are!zero).!(g) Holding!the!variables!head!length!and!neck!circumference!fixed!(it!may!be!easier!to!state!“for!bears!with!the!same!length!and!neck!measurements”),!the!estimated!weight!of!a!bear!increases!by!9.36386!pounds!as!the!chest!circumference!of!a!bear!increases!by!one!inch.!(h) )ˆ(ˆ2)975.0(12ββSEtmn −−±is!a!95%!confidence!interval!for!the!regression!coefficient!β2!associated!with!length.!!Plugging!the!values!in,!we!have!–0.10288!±!2.009(0.90867),!or!a!CI!of!(–!1.93,!1.72)!inches.!!!Note!that!the!CI!provides!evidence!that!the!partial!regression!coefficient!β2!for!head!length!is!not!significantly!different!from!0,!since!0!is!contained!in!the!95%!CI!for!β2!(i.e.,!zero!is!a!plausible!value!for!β2).!!Thus,!after!including!neck!and!chest!circumference!in!the!model!for!the!mean!weight!of!bears,!it!appears!that!the!variable!X2!(head!length)!does!not!linearly!affect!the!mean!weight!and!can!be!dropped!from!the!model.!!!(i) R2!=!0.94!(coefficient!of!determination)!provides!the!proportion!of!variation!in!weight!that!is!explained!by!this!regression.!(j) µˆ(Y'|!X1!=!35,!X2!=!60,!X3!=!24)!=!U!264.78856!+!9.36386(35)!–!0.10288(60)!+!5.82641(24)!=!196.6!lbs.!(k) 95%!CI!for!µ(Y|X1!=!35,!X2!=!60,!X3!=!24)!is!(177.9!to!215.3)!lbs!(l) Pred(Y'|!X1!=!35,!X2!=!60,!X3!=!24)!=!U!264.78856!+!9.36386(35)!–!0.10288(60)!+!5.82641(24)!=!196.6!lbs.!(m) 95%!prediction!interval!for!the!weight!Y!of!a!randomly!captured!bear!with!X1!=!35,!X2!=!6,!!X3!=!24!measurements!(130.5,!262.7)!lbs!!(n) The!plots!of!the!data!against!the!explanatory!variables!suggest!a!slight!increase!in!variances!and!a!slight!nonlinearity.!!These!features!are!more!visible!in!the!residual!plots,!which!remove!the!effects!of!all!of!the!explanatory!variables.!!The!normal!plot!indicates!some!deviation!from!normality,!particularly!due!to!two!outlying!residuals!(which!should!be!investigated).!!!2) Problem 2 (three corn varieties and their response to fertilizer)(a) The population regression model isµ(Y |X1, X2, X3, X4, X5) = β0+ β1X1+ β2X2+ β3X3+ β4X4+ β5X5,where Y = yield (kg), X1= fertilzer amount, X2= 1 if variety A is planted in the plot and = 0otherwise, X3= 0 if variety B is planted in the plot and = 0 otherwise, X4= X1X2, and X5= X1X3.(b) Test the null hypothesis that says that the mean µ(Y |X1, X2, X3, X4, X5) of Y does not dependon X1, X2, X3, X4, and X5.Test H0: β1= β2= β3= β4= β5= 0 vs Ha: at least one coefficient among β1, β2, β3, β4, β5isnot zeroUse F = 316.20 from multiple regression ANOVA table, df1= p − 1 = m = 5 & df2= n − p =12 − 6 = 6Since p-value< 0.0001, we have strong evidence that not all regression coefficients β1, β2, β3, β4, β5are zero. So, some collection of variables Ximay be important for describing the mean yield of cornin a linear way.(c) The least squares regression line that estimaes the model in (a) is130 + 1.38X1− 2.5X2+ 5.5X3− 0.12X4− 0.78X5(d) Models for each variety:• Variety A: X1, X2= 1, X3= 0, X4= X1, X5= 0 ⇒µ(Y |X1, X2, X3, X4, X5) = (β0+ β2) + (β1+ β4)X1• Variety B: X1, X2= 0, X3= 1, X4= 0, X5= X1⇒µ(Y |X1, X2, X3, X4, X5) = (β0+ β3) + (β1+ β5)X1• Variety C: X1, X2= X3= X4= X5= 0 ⇒µ(Y |X1, X2, X3, X4, X5) = β0+ β1X1So under this multiple regression model with indicator variables X2, X3and interaction variablesX4, X5, we see that the mean yield µ(Y ) of corn is linearly related to fertilizer amount X1for eachvariety of corn A, B, C. In addition, each variety could possibly have a different regression line dueto different intercepts and different slopes (the slope being the coefficient in front of fertilizer amountX1).1Keep in mind too that population regression line describing the mean yield µ(Y |X1, X2, X3, X4, X5)and its connection to variety type and fertilizer amount is something we don’t really know; namely, wedon’t know any of the coefficients βiabove. The mean yield µ(Y |X1, X2, X3, X4, X5) represents theaverage yield for all potential plots planted with a corn variety A, B or C and a fertilizer amountX1(under similar environmental conditions). We don’t know about the potential yields of all possibleplots, since we just have a sample collected in one experiment. But based on this data, we can estimatethese βi’s and, then using our estimatesˆβi, we can determine if the data suggest some βiare zero ornot.(e-g) The question of interest is whether the effect of fertilizer amount X1on the mean yield µ(Y )of corn is the same for all three corn varieties. Since for each variety, there is a regression line thatlinearly relates amount X1to the mean yield µ(Y ), the effect of amount X1is really determined bythe slopes (β1+ β4), (β1+ β5) and β1for each respective variety. In the following, we can examineif the regression lines for each variety have the same slopes (i.e., are the slopes (β1+ β4), (β1+ β5)and β1all equal? In other words does β4= β5= 0?) If some varieties have the same slopes in theirregression lines, then we would find that the effect of the amount of fertilizer is really the same of thesevarieties. Thus, for this problem, we need to test H0: β4= β5= 0 vs. Ha: either β46= 0 or β56= 0or both (i.e., not H0).Under the null hypothesis, we have a reduced/simplified model to describe the mean yield. Underthe alternative hypothesis, we have a full (more parameters) model to describe the mean yield. Notethat the intercept has nothing to do with fertilizer. Because we have